# Thread: Integration Problem

1. ## Integration Problem

Hello, I'm new here and I have a problem that I can't solve. I thank you in advance for ANY help you can give me. I know things are written to scale, but I couldn't find where to access the symbols. Thank you so much!

∫ x - 1
----------dx
2
x - 4x + 5

this bottom x is squared and the -- lines are the divided by lines. Then the formula is multiplied by dx. Thanks!

Cindi

2. Originally Posted by Cindi24
Hello, I'm new here and I have a problem that I can't solve. I thank you in advance for ANY help you can give me. I know things are written to scale, but I couldn't find where to access the symbols. Thank you so much!

∫ x - 1
----------dx
2
x - 4x + 5

this bottom x is squared and the -- lines are the divided by lines. Then the formula is multiplied by dx. Thanks!

Cindi
note that $\int \frac {x - 1}{x^2 - 4x + 5}~dx = \frac 12 \int \frac {2x - 2}{x^2 - 4x + 5}~dx$ $= \frac 12 \int \frac {2x - 4 + 2}{x^2 - 4x + 5}~dx = \frac 12 \int \frac {2x - 4}{x^2 - 4x + 5}~dx + \int \frac 1{x^2 - 4x + 5}~dx$

for the first integral, do a substitution of $u = x^2 - 4x + 5$, for the second, complete the square in the denominator, and try to get the arctangent integral.

can you continue?

3. Originally Posted by o_O
$\frac{x-1}{x^{2}-4x+5} = \frac{x-1}{(x+5)(x-1)} = \frac{1}{x+5}$

Easier to deal with?
that's not true. if you expand the denominator of what you wrote, you get $x^2 + 4x - 5$

4. oh, okay! I think so. Thank you!

5. Ah whoops my bad. Don't know what I was thinking =S

6. Originally Posted by o_O
Ah whoops my bad. Don't know what I was thinking =S
hehe, i thought the same thing at first, but then i realized the factorization doesn't work, so i had to do it the hard way